Java 无效列类型:未为类oracle.jdbc.driver.T4CRowidAccessor实现getInt
我想在oracle数据库11g中发布数据,我使用输入字段输入数据。数据输入到数据库,但只有id coulmn和另一个字符串coulmn获得正确的数据,但另一个coulmn获得null或0,因此我无法解决此问题,在运行时我出现此错误Java 无效列类型:未为类oracle.jdbc.driver.T4CRowidAccessor实现getInt,java,jdbc,oracle11g,Java,Jdbc,Oracle11g,我想在oracle数据库11g中发布数据,我使用输入字段输入数据。数据输入到数据库,但只有id coulmn和另一个字符串coulmn获得正确的数据,但另一个coulmn获得null或0,因此我无法解决此问题,在运行时我出现此错误 列类型无效:未为类oracle.jdbc.driver.T4CRowidAccessor实现getInt 这是我的密码 public Client newClient(Client client){ try { con = DBConn
列类型无效:未为类oracle.jdbc.driver.T4CRowidAccessor实现getInt
这是我的密码
public Client newClient(Client client){
try {
con = DBConnection.getConnection(driver, url, name, pass);
pstmt = con.prepareStatement("INSERT INTO CLIENT (CID, FIRSTNAME, LASTNAME, CAREER, CSALARY) VALUES (CLIENT_ID.nextval, ?, ?, ?, ?)", Statement.RETURN_GENERATED_KEYS);
pstmt.setString(1, client.getFirstName());
pstmt.setString(2, client.getLastName());
pstmt.setString(3, client.getCareer());
pstmt.setInt(4, client.getcSalary());
pstmt.executeUpdate();
rs = pstmt.getGeneratedKeys();
rs.next();
Integer id = rs.getInt(1);
client.setcId(id);
}catch(Exception ex){
ex.printStackTrace();
return null;
}finally{
try{ rs.close(); }catch (Exception e){}
try{ pstmt.close();}catch (Exception e){}
try{ con.close();}catch (Exception e){}
}//finally
return client;
}
}
}
这是我的SQL
CREATE TABLE "AHMAD"."CLIENT"
( "CID" NUMBER(*,0) NOT NULL ENABLE,
"FIRSTNAME" VARCHAR2(255 BYTE),
"LASTNAME" VARCHAR2(255 BYTE),
"CAREER" VARCHAR2(255 BYTE),
"CSALARY" NUMBER,
CONSTRAINT "CLIENT_PK" PRIMARY KEY ("CID")
USING INDEX PCTFREE 10 INITRANS 2 MAXTRANS 255
STORAGE(INITIAL 65536 NEXT 1048576 MINEXTENTS 1 MAXEXTENTS 2147483645
PCTINCREASE 0 FREELISTS 1 FREELIST GROUPS 1 BUFFER_POOL DEFAULT FLASH_CACHE DEFAULT CELL_FLASH_CACHE DEFAULT)
TABLESPACE "USERS" ENABLE
) SEGMENT CREATION IMMEDIATE
PCTFREE 10 PCTUSED 40 INITRANS 1 MAXTRANS 255 NOCOMPRESS LOGGING
STORAGE(INITIAL 65536 NEXT 1048576 MINEXTENTS 1 MAXEXTENTS 2147483645
PCTINCREASE 0 FREELISTS 1 FREELIST GROUPS 1 BUFFER_POOL DEFAULT FLASH_CACHE DEFAULT CELL_FLASH_CACHE DEFAULT)
TABLESPACE "USERS" ;
请帮助我将数据插入到表中,但不完整,我有此运行时错误
java.sql.SQLException: Invalid column type: getInt not implemented for class oracle.jdbc.driver.T4CRowidAccessor
at oracle.jdbc.driver.DatabaseError.throwSqlException(DatabaseError.java:112)
at oracle.jdbc.driver.DatabaseError.throwSqlException(DatabaseError.java:146)
at oracle.jdbc.driver.Accessor.unimpl(Accessor.java:358)
at oracle.jdbc.driver.Accessor.getInt(Accessor.java:468)
at oracle.jdbc.driver.OracleReturnResultSet.getInt(OracleReturnResultSet.java:265)
at com.rest.client.ClientDAO.newClient(ClientDAO.java:99)
at com.rest.client.ClientResourcs.newClient(ClientResourcs.java:36)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
at java.lang.reflect.Method.invoke(Unknown Source)
at com.sun.jersey.spi.container.JavaMethodInvokerFactory$1.invoke(JavaMethodInvokerFactory.jav a:60)
at com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$TypeOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:185)
at com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:75)
at com.sun.jersey.server.impl.uri.rules.HttpMethodRule.accept(HttpMethodRule.java:288)
at com.sun.jersey.server.impl.uri.rules.ResourceClassRule.accept(ResourceClassRule.java:108)
at com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
at com.sun.jersey.server.impl.uri.rules.RootResourceClassesRule.accept(RootResourceClassesRule.java:84)
at com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1469)
at com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1400)
at com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1349)
at com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1339)
at com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:416)
at com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:537)
at com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:708)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:305)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:222)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:123)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:472)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:171)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:99)
at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:936)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:407)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1004)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:589)
at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:310)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)
Oracle JDBC驱动程序在使用
语句时不返回生成的id。返回生成的\u KEYS
,因此生成的KEYSResultSet
的第一列不是int
,而是返回插入行的ROWID
(请参见)以便您自己查询该行。据我所知,这是因为Oracle没有标识类型,所以它实际上不知道生成的密钥是哪个字段
相反,您应该能够使用or显式指定要返回的列。就像Mark建议的那样,我只是遇到了同样的问题并找到了解决方案:
String generatedColumns[] = { "ID" };
stmt = con.prepareStatement(query, generatedColumns);
执行查询后,执行以下操作以获取插入的ID
rs = stmt.getGeneratedKeys();
rs.next();
rs.getInt(1);
不知道这个问题是否得到解决,但我遇到了相同的问题,通过在实体类中将整数替换为int(在您的情况下是客户端类)解决了这个问题数据插入到表中,id是自动递增的,但整行就像1003 null Dev 0,所以请原谅我是初学者,让错误再次读取我的答案。问题是,您认为getGeneratedKeys的第一列包含插入行的
CID
列的值,但它没有。它包含rowid,这是Oracle定义的一个特殊伪列11g@AchJ你说得对,我最近在Oracle上使用了prepareStatement(String sql,String[]columnNames)
,所以我删除了这个警告。谢谢!有一段时间我遇到了这个错误,但在插入后我从不需要使用任何id。。但现在我有了这个需要,这个问题必须解决!注意:为什么oracle如此过时??!
rs = stmt.getGeneratedKeys();
rs.next();
rs.getInt(1);